Question

Find the direction cosines of the sides of the triangle whose vertices are $(3,5,-4)$ and $(-1,1,2)$ and $(-5,-5,-2)$. What type triangle is it?

Answer 1

$A(3,5,-4);B(-1,1,2);C(-5,-5,-2)$ are vertices of $△ABC$

$\therefore \overrightarrow{AB}=(-1-3,1-5,2-(-4\left)\right)$

$\therefore \overrightarrow{BC}=(-5-(-1),-5-1,-2-2)$

$\therefore \overrightarrow{CA}=(3-(-5),5-(-5),-4-(-2\left)\right)$

$\therefore \overrightarrow{AB}=(-4,-4,6)$

$\therefore \overrightarrow{BC}=(-4,-6,-4)$

$\therefore \overrightarrow{CA}=(8,10,-2)$

$\therefore$ direction cosines of $AB=\frac{1(-4,-4,6)}{\sqrt{(-4{)}^2+(-4{)}^2+62}}$

$=(\frac{-4}{\sqrt{68}},\frac{-4}{\sqrt{68}},\frac{-4}{\sqrt{68}})$

$\therefore$ direction cosines of $BC=\frac{(-4,-6,-4)}{\sqrt{(-4{)}^2+(-6{)}^2+(-4{)}^2}}$

$=(\frac{-4}{\sqrt{6}8},\frac{-6}{\sqrt{68}},\frac{-4}{\sqrt{68}})$

$\therefore$ direction cosines of $BC=\frac{(-4,-6,-4)}{\sqrt{(-4{)}^2+(-6{)}^2+(4{)}^2}}$

$=(\frac{-4}{\sqrt{6}8},\frac{-6}{\sqrt{68}},\frac{-4}{\sqrt{68}})$

$\therefore$ direction cosines of $CA=\left(\frac{8,10,-2}{\sqrt{8^2+{10}^2+(-2{)}^2}}\right)$

$=(\frac{8}{\sqrt{168}},\frac{10}{\sqrt{168}},\frac{-2}{\sqrt{168}})$